Question

(Requires calculus) Suppose that \(f\left( x \right) = {e^x}\) and \(g\left( x \right) = {e^{cx}}\) where c is a constant. Use mathematical induction together with the chain rule and the fact that \(f'\left( x \right) = {e^x}\) to prove that \({g^{\left( n \right)}} = {c^n}{e^{cx}}\) whenever \(n\) is a positive integer.

Short answer

It is proven that \({g^{\left( n \right)}}\left( x \right) = {c^n}{e^{cx}}\) whenever \(n\) is a positive integer.

Step-by-step solution

Principle of Mathematical Induction

To prove that\(P\left( n \right)\)is true for all positive integers\(n\), where \(P\left( n \right)\)is a propositional function, we complete two steps:

Basis Step:

We verify that \(P\left( 1 \right)\)is true.

Inductive Step:

We show that the conditional statement\(P\left( k \right) \to P\left( {k + 1} \right)\) is true for all positive integers k.

Proving the basis step

Let\(P\left( n \right)\):\({g^{\left( n \right)}}\left( x \right) = {c^n}{e^{cx}}\)

In the basis step, we need to prove that\(P\left( 1 \right)\)is true

For finding statement\(P\left( 1 \right)\)substituting\(1\)for\(n\)in the statement

\({g^{(1)}}\left( x \right) = c{e^x}\)

Given function is

\(g\left( x \right) = {e^{cx}}\)

Differentiating with respect to x

\(g'\left( x \right) = \frac{d}{{dx}}\left( {{e^{cx}}} \right)\)

Using chain rule of differentiation.

\(\begin{aligned}{c}g'\left( x \right) &= \left( {{e^{cx}}} \right)\frac{d}{{dx}}\left( {cx} \right)\\ &= c{e^{cx}}\end{aligned}\)

From the above, we can see that the statement \(P\left( 1 \right)\) is true this is also known as the basis step of the proof.

Proving the Inductive step

In the inductive step, we need to prove that, if\(P\left( k \right)\)is true, then\(P\left( {k + 1} \right)\)is also true.

That is,

\(P\left( k \right) \to P\left( {k + 1} \right)\)is true for all positive integers k.

In the inductive hypothesis, we assume that\(P\left( k \right)\)is true for any arbitrary positive integer\(k\)

That is

\({g^{\left( k \right)}}\left( x \right) = {c^k}{e^{cx}}\)

Differentiating again with respect to x

\({g^{\left( {k + 1} \right)}}\left( x \right) = \frac{d}{{dx}}\left( {{c^k}{e^{cx}}} \right)\)

Again, using the chain rule of differentiation

\(\begin{aligned}{c}{g^{\left( {k + 1} \right)}}\left( x \right) &= \frac{d}{{dx}}\left( {{c^k}{e^{cx}}} \right)\\ &= {c^k}{e^{cx}}\frac{d}{{dx}}\left( {cx} \right)\\ &= {c^k}{e^{cx}}\left( c \right)\\ &= {c^{k + 1}}{e^{cx}}\end{aligned}\).

From the above, we can see that\(P\left( {k + 1} \right)\)is also true

Hence,\(P\left( {k + 1} \right)\)is true under the assumption that\(P\left( k \right)\)is true. This

completes the inductive step.

Hence It is proved \({g^{\left( n \right)}}\left( x \right) = {c^n}{e^{cx}}\) whenever \(n\) is a positive integer.